- •1.2 THEORETICAL INTRODUCTION
- •1.3. WORK SEQUENCE
- •1.4 APPENDIX
- •1.5 TEST QUESTIONS
- •1.6 REFERENCES
- •2 LABORATORY PRACTICUM: EQUILIBRIUM OF HOMOGENEOUS CHEMICAL SYSTEMS
- •2.1 THEORETICAL INTRODUCTION
- •3.1. THEORETICAL INTRODUCTION
- •2.2. TEST QUESTIONS
- •2.3. REFERENCES
- •Limited Mutual Solubility of Liquids
- •Distribution of the Third Component between Two Immiscible Liquids
- •The used research method is titration.
- •Experiment Procedure
- •The used research method is titration.
- •The used research method is titration.
- •Reagents and materials: a 0.05 M (0.1 N) iodine solution in carbon tetrachloride, a 0.001 M sodium thiosulphate (Na2S2O3) solution, and a 1% freshly prepared aqueous solution of starch.
- •3.2. TEST QUESTIONS
- •3.3. TASKS FOR SELF-STUDY
- •=const,
- •Solution. Let us calculate the K values by the equation,
- •Taking a logarithm of both parts of the expression, one finds that
- •b) The following equation should be used for the case of five consecutive extractions:
- •Problems
- •3.4. REFERENCES
- •4.1. THEORETICAL INTRODUCTION
- •4.2. TEST QUESTIONS
- •LABORATORY EXERCISE 10.
- •4.3. APPENDIX
- •4.4. TEST QUESTIONS
- •4.5 REFERENCES
- •1. Explain the term "molecularity of a chemical reaction". Can the molecularity be greater or smaller than the reaction order?
- •4. Upon studying the kinetics of a chemical reaction, the kinetic curves with different concentrations of reagents have been obtained. Which of the methods of determination of the reaction order is most effective in this case?
- •w = k[HCrO4–][3HSO3–]2[H+].
- •Why is the rate of this reaction not proportional to the number of ions of each sort in accordance with the stoichiometric coefficients in the chemical equation?
- •5.2. KINETICS OF COMPLEX CHEMICAL REACTIONS
- •Task 3
- •5.3. REFERENCES
- •6. INDIVIDUAL ASSIGNMENTS. ELECTROLYTE SOLUTIONS
-lgK=0.66; K=0.219.
The partition law for the specified system is expressed by the equation,
C 1.66
C1 2 = 0.219 .
Answer: The distribution coefficient of salicylic acid between water and benzene is 0.219, and the exponent on the concentration value for an aqueous solution equals 1.66.
Sample problem № 6. The distribution coefficient of iodine between water and carbon disulphide equals 0.0017. An aqueous solution of iodine containing 10–4 kg of iodine in 10–4 m3 of a solution was shaken with carbon disulphide. What amount of iodine remains in an aqueous solution, if (a) 10–4 m3 of an aqueous solution were shaken with 5·10–6 m3 of carbon disulphide, and (b) 10–4 m3 of an aqueous solution were shaken with five
separate portions of carbon disulphide, under the condition that the volume of each portion is 10–6 m3?
Solution.
a) In the first case only one extraction was performed. Using the
equation, |
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g1 = g0 |
KV1 |
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one can calculate the amount of iodine remained in an aqueous solution:
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0.0017 10−4 |
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0.0017 10−4 +5 10−6 |
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The amount of separated iodine is 0.967·10-4 kg.
b) The following equation should be used for the case of five consecutive extractions:
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gm = g0 |
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g5 |
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KV +V |
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g5 =10 |
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= 6.5·10 kg. |
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